Question : If $x=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$ and $y=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$, then the value of $x^{3} + y^{3}$ is:
Option 1: 950
Option 2: 730
Option 3: 650
Option 4: 970
Correct Answer: 970
Solution : $x=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}=\frac{(\sqrt{3}-\sqrt{2})×(\sqrt{3}-\sqrt{2})}{(\sqrt{3}+\sqrt{2})×(\sqrt{3}-\sqrt{2})}=(\sqrt{3}-\sqrt{2})^2$ = $5-2\sqrt6$ $y=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}=\frac{(\sqrt{3}+\sqrt{2})×(\sqrt{3}+\sqrt{2})}{(\sqrt{3}-\sqrt{2})×(\sqrt{3}+\sqrt{2})}=(\sqrt{3}+\sqrt{2})^2$ = $5+2\sqrt6$ $⇒x+y=5-2\sqrt6+5+2\sqrt6 =10$ Also, $xy=(5-2\sqrt6)(5+2\sqrt6)=25-24=1$ $\therefore x^3+y^3=(x+y)^{3} -3xy(x+y)$ $=10^3-3\times 1 \times 10$ $=1000-30$ $= 970$ Hence, the correct answer is 970.
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